3
$\begingroup$

Setup: Consider the Feynman propagator as it appears in Peskin and Schroeder eq. 2.59:

\begin{align} D_F(x-y)\equiv i\int_{\mathbb{R}^4} \frac{d^4p}{(2\pi)^4}\frac{e^{ip\cdot(x-y)}}{p^2-m^2+i\epsilon}, \end{align} where $p,x,y$ denote 4-vectors; $p^2\equiv \eta_{\mu\nu}p^\mu p^\nu$, $p\cdot (x-y)\equiv \eta_{\mu\nu}p^\mu (x-y)^\nu $; the sign convention here is $\text{diag}(\eta)=(+1,-1,-1,-1)$; and the $i\epsilon$ has been inserted for convergence.

The following "formal" manipulations would seem to imply $i\int d^4x D_F(x-y)=1/m^2$.

\begin{align} i\int_{\mathbb{R}^4}d^4x D_F(x-y)&=-\int_{\mathbb{R}^4} d^4x \int_{\mathbb{R}^4}\frac{d^4p}{(2\pi)^4}\frac{e^{ip\cdot(x-y)}}{p^2-m^2+i\epsilon}\\ &=- \int_{\mathbb{R}^4}d^4p\frac{e^{-ip\cdot y}}{p^2-m^2+i\epsilon}\int_{\mathbb{R}^4} \frac{d^4x}{(2\pi)^4}e^{ip\cdot x}\\ &=- \int_{\mathbb{R}^4}d^4p\frac{e^{-ip\cdot y}}{p^2-m^2+i\epsilon} \delta^{(4)}(p)\\ &=\frac{1}{m^2}, \tag{*}\label{*} \end{align} where I used a standard $\delta$-identity in going from the 2nd to the 3rd line, and took the $\epsilon\to 0$ limit after performing the $p$-integral.

Question(s) These manipulations are "formal" in the sense that I have interchanged the order of integration without justification. I suppose, in general, one would want to apply something akin to Fubini's theorem. However, I am not sure to what extent the regularization implicit in the Feynman propagator might alter such an argument.

Can one justify these formal manipulations? To what extent is $\eqref{*}$ a true result and is there any physical intuition for it?

$\endgroup$
  • 1
    $\begingroup$ Aren't you just picking the $p = 0$ component of the propagator by integrating over all $x$? $\endgroup$ – Darkseid Dec 14 '17 at 20:35
  • $\begingroup$ Could you elaborate? My formal manipulations indicate only the $p=0$ part contributes, but I don't know to what extent these manipulations should be trusted. Nor is it intuitive to me why only the $p=0$ part should contribute to the $x$-integral. $\endgroup$ – user143410 Dec 14 '17 at 20:43
  • 1
    $\begingroup$ @user143410 You understand the relationship between Taylor series and function moments, and how they swap places under Fourier transformation, right? So, evaluating the Fourier transformed function at $p=0$ is the same as evaluating the integral, and evaluating the original function at $x=0$ is the same as taking the zeroth moment of the Fourier transformed function. $\endgroup$ – Sean E. Lake Dec 14 '17 at 23:33
  • $\begingroup$ @user143410 this is what $p = 0$ component is, a sum over all $x$. This might be considered a part of definition of the Fourier Transform. $\endgroup$ – Darkseid Dec 15 '17 at 9:08
  • 1
    $\begingroup$ @user143410 Nope. I mean if you take the formula for the $n^{\mathrm{th}}$ moment $$M_n = \int x^n f(x) \operatorname{d}x $$ and drop in $f(x) = \int \tilde{f}(k) \frac{\mathrm{e}^{ikx}}{\sqrt{2\pi}} \operatorname{d}k$ you'll find that $$M_n = \left.(-i)^n\frac{\partial^n \tilde{f}}{\partial k^n}\right|_{k=0}.$$ $\endgroup$ – Sean E. Lake Dec 15 '17 at 18:08
5
$\begingroup$

All of this becomes mathematically rigorous within the theory of temperate Schwartz distributions and their Fourier transforms. Put $y=0$ for simplicity. Then $D_{F}(x)$ is in $S'(\mathbb{R}^4)$. Now you are taking its Fourier transform. No problem. Then you are evaluating it at zero momentum. That needs care because the Fourier transform (which is also the propagator in momentum space) is a distribution too. Fortunately, zero is not in the singular support of this distribution, so pointwise evaluation is legit and gives the $1/m^2$. Your computation simply amounts to the Fourier inversion theorem for a temperate distribution. To learn about distributions see the old book by Laurent Schwartz "Mathematics for the Physical Sciences", or the book by Strichartz "A Guide to Distribution Theory and Fourier Transforms".

A lower tech approach (but with the theory of distributions lurking in the background) is to do Fubini etc. with the propagator $$ \frac{e^{-\epsilon_1 p^2}}{p^2-m^2+i\epsilon_2} $$ and then take the limits $\epsilon_1,\epsilon_2\rightarrow 0^{+}$. Careful: the $p^2$ in the exponential is with Euclidean signature. This brings you from $S'(\mathbb{R}^4)$ to $S(\mathbb{R}^4)$ and basically proving the Fourier inversion formula in the latter space. In fact this proof usually requires introducing another $e^{-\epsilon_3 x^2}$, again with Euclidean signature. Good luck with all the epsilons.

PS: I said the theory of distributions is lurking in the background because of the following two facts. The space $S(\mathbb{R}^d)$ of rapidly decaying functions (for which classical tools like Fubini etc. work well) is dense in the space of distributions $S'(\mathbb{R}^d)$. The second fact is that operations on distributions like taking the Fourier transform are defined as the unique continuous extensions of their classical versions on $S(\mathbb{R}^d)$. So putting all the epsilons amounts to doing distributional analysis classically, i.e., at the level of a first graduate course in analysis, measure theory etc.

$\endgroup$
1
$\begingroup$

Only formally, in the same sense that $$\int_{-\infty}^\infty \operatorname{d}x\, \frac{\operatorname{e}^{ikx}}{2\pi} = \delta(k).$$ Why? You can check that if you integrate the Feynman propagator in position space over all space you'll get $$\int \operatorname{d}^3y\, D_F(x-y) = \frac{1}{2m}\sin(m|x^0-y^0|) - \frac{i}{m} \cos(m[x^0-y^0]), $$ where I may have screwed up the scale in front of the cosine term, but the one in front of the sine term is right. Getting that sine term right is an essential part of satisfying the boundary conditions in the defining equation for the propagator $$\Box D_F(x-y) + m^2 D_F(x-y) = \delta^4(x-y).$$ Just integrate both sides of the defining equation over all space to see what I'm talking about.

Doing the time integral on the sine and cosine terms, you'll find that it converges in the mean to \begin{align} \int \operatorname{d}x^0 \left[\frac{1}{2m}\sin(m|x^0-y^0|) - \frac{i}{m} \cos(m[x^0-y^0])\right] &= \frac{1}{m^2}\left[-\cos(mu) - 2i \sin(mu)\right]_{u=0}^{u_\mathrm{max}} \\ & = \frac{1}{m^2} - \frac{\cos(mu_{\mathrm{max}}) +2i\sin(mu_{\mathrm{max}})}{m^2} \\ &\rightarrow \frac{1}{m^2}\ \mathrm{in\ the\ mean.} \end{align}

$\endgroup$
  • $\begingroup$ I like your position space evaluation which makes explicit the kind of "approximation"/averaging being made with the formal manipulations. I wonder if the Minkowski propagator is equally or less convergent in this sense than the Euclidean one. Naively, the Euclidean propagator should contain divergences only at the origin (b/c $|x_1-x_2|=0$ only when $x_1=x_2$), while the Minkowski one might be divergent along the entire lightcone (b/c $|x_1-x_2|=0$ on the lightcone). $\endgroup$ – user143410 Dec 15 '17 at 17:20
  • 1
    $\begingroup$ @user143410 The Euclidean propagator is completely convergent - it is just a modified Bessel function of the second kind in every direction, so exponentially damped. When you integrate off all but one direction of that one, you get $$\sqrt{\frac{|x-x'|}{2\pi m}}K_{-1/2}(m|x-x'|)=\frac{1}{2m}\exp(-m|x-x'|).$$ Minkowski is only finite for $d=2$, otherwise you have a divergence as you approach the light cone from within ($d$ not even), or a delta function on it ($d$ even). $\endgroup$ – Sean E. Lake Dec 15 '17 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.